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Calculus Level 3

If S n = ∑ r = 1 n 2 r + 1 r 4 + 2 r 3 + r 2 S_{n}=\sum_{r=1}^n\frac{2r+1}{r^{4}+2r^{3}+r^{2}} then S 20 S_{20} is equal to n n + 1 \frac{n}{n+1} then find n + 1 \sqrt{n+1}


The answer is 21.

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Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Chew-Seong Cheong
Apr 12, 2015

Same answer as Shivam Jadhav . Maybe better LaTex.

S k = ∑ r = 1 k 2 r + 1 r 4 + 2 r 3 + r 2 = ∑ r = 1 k 2 r + 1 r 2 ( r + 1 ) 2 = ∑ r = 1 k ( 1 r 2 − 1 ( r + 1 ) 2 ) = 1 1 − 1 ( k + 1 ) 2 = k ( k + 2 ) ( k + 1 ) 2 S_k = \displaystyle \sum_{r=1}^k {\dfrac {2r+1}{r^4+2r^3+r^2}} = \sum_{r=1}^k {\dfrac {2r+1}{r^2(r+1)^2}} = \sum_{r=1}^k {\left( \dfrac {1}{r^2} - \dfrac {1}{(r+1)^2} \right)} \\ \quad = \dfrac {1}{1} - \dfrac {1}{(k+1)^2} = \dfrac {k(k+2)}{(k+1)^2}

⇒ S 20 = 20 ( 22 ) 2 1 2 = 440 441 ⇒ n = 440 ⇒ 440 + 1 = 21 \Rightarrow S_{20} = \dfrac {20(22)}{21^2} = \dfrac {440}{441} \quad \Rightarrow n = 440 \quad \Rightarrow \sqrt{440+1} = \boxed{21}

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